For the square of AB is equal to the squares of AF, FB. But (Eucl. ii. 5.) the square of FB is equal to the rectangle BE, EG together with the square of EF, i. e. to the rectangle CE, ED together with the square of EF; .. the square of AB is equal to the squares of AF, FE together with the rectangle CE, ED, i. e. to the square of AE together with the rectangle CE, ED.

(14.) A straight line drawn from the concourse of two tangents to the concave circumference of a circle is divided harmonically by the convex circumference and the chord which joins the points of contact.

Let AB, AC touch the circle ADC, and AGE cut it. Join BC; then will

AE AG :: EF : FG.

On EG as diameter describe a

D

E

H

B

circle EHG, and through F draw HFI perpendicular to EG. Join AH. Then the rectangle BF, FC is equal to the rectangle EF, FG, i. e. to the square of HF, or the rectangle HF, FI; and .. the are in the circumference of a circle.

points H, B, I, C And since the And since the square

AF, FH, or to the

of AH is equal to the squares of square of AF and the rectangle BF, FC, i. e. to the squares of AK, KF, together with the rectangle BF, FC, i. e. to the squares of AK, KB, or to the square of AB; .. AH=AB; and since the square of AH is equal to the rectangle EA, AG, AH is a tangent at H. And since

EG is a diameter, (ii. 42.)

AE AG: EF: GF.

(15.) If from the extremities of any chord in a circle straight lines be drawn to any point in the circumference meeting a diameter perpendicular to the chord; the rectangle contained by the distances of their points of intersection from the centre is equal to the square described upon the radius.

From A and B, the extremities of the chord AB, let AC, BC be drawn to any point C in the circumference; and let them meet a diameter perpendicular to AB in D and E. Take O

B

F/E

the centre; the rectangle DO, OE is equal to the square described on the radius.

Draw the diameter BOG. Join CG, CO. Since the angle OCB is equal to OBC, and BGC to FAD, and that CBG, and BGC together are equal to a right angle; .. OCE and FAD together are equal to a right angle, and.. to FAD and ADF together; hence OCE is equal to ADO, .. the triangles COD, COE are equiangular,

and DO OC :: OC: OE.

.. the rectangle DO, OE is equal to the square of OC.

(16.) If from any point in the base or base produced, of the segment of a circle, a line be drawn making therewith an angle equal to the angle in the segment, and from the extremity of the base any line be drawn to the former, and cutting the circumference; the rectangle contained by this line and the part of it within the sement is always of the same magnitude.

M M

B

Let ABC be a segment of a circle on the base AC; and from any point D, let DE be drawn, making with AC an angle equal to the angle in the ment, and meeting any line AB drawn from the extremity A; the rectangle EA, AB is of invariable magnitude.

seg

C D

Since the angle ADE is equal to ABC, and the angle at A common to the triangles ADE, ABC, the triangles are .. similar; whence

AD AE AB : AC,

and the rectangle AE, AB is equal to the rectangle AD, AC, which is invariable.

(17.) To determine the locus of the extremities of any number of straight lines drawn from a given point, so that the rectangle contained by each and a segment cut off from each by a line given in position may be equal to a given rectangle.

Let A be the given point, and DE the line given in position. Draw AGF perpendicular to DE; and take AF such that the rectangle AG, AF may be equal to the given rectangle; and on AF as diameter describe a circle; it will be the locus required.

B

E

G

Draw any line AC; and join FC. The triangles ABG, AFC being similar,

AF : AC :: AB : AG,

..the rectangle AC, AB is equal to the rectangle AF, AG, i. e. to the given rectangle. And the same may be proved of any other line drawn from A to the circumference, which .. is the locus.

(18.) If from a given point two straight lines be drawn, containing a given angle, and such that their rectangle may be equal to a given rectilineal figure, and one of them be terminated by a straight line given in position; to determine the locus of the extremity of the other.

Let A be the given point, and BC the line given in position. From A draw AD perpendicular to BC; and draw AE, making with it the angle DAE equal to the given angle; and make AE of such a magnitude that the rectangle AD, AE may be equal to the given figure. On AE as diameter describe a circle AFE; it will be the locus required.

B D

Draw any other line AB, and AF making with it the angle FAB equal to the given angle; join FE. Then the triangles ABD, AFE, being equiangular,

AB: AD :: AE : AF,

whence the rectangle AB, AF is equal to the rectangle AD, AE, i. e. to the given figure; and the same may be proved, of any other two lines, similarly drawn from A.

(19.) If from the vertical angle of a triangle two lines be drawn to the base making equal angles with the adjacent sides; the squares of those sides will be proportional to the rectangles contained by the adjacent segments of the base.

+

Let AD, AE be drawn from the vertical angle A making equal angles BAD, EAC with the adjacent sides; then will AB: AC2 :: BD× BE : CD × CE.

About the triangle ADE describe a circle, cutting AB, AC (produced if necessary) in G and F. Join FG. Then (Eucl. iii. 26.) the arcs GD, FE are equal,.. (ii. 1. Cor.) FG is parallel to BC;

and AB AC2 :: AB× BG : AC× CF

:

:: BD× BE: CDx CE (Eucl. iii. 36.).

(20.) If a line placed in one circle be made the diameter of a second, the circumference of the latter passing through the centre of the former; and

any chord in the former circle be drawn through this diameter perpendicularly; the rectangle contained by the segments made by the circumference of the latter circle will be equal to that contained by the whole diameter and a mean proportional between its segments.

Let a line AC, placed in the circle ADC, be the diameter of the circle ABC, whose circumference passes through the centre of ADC. Through any point B let a line DBE be drawn perpendicular to AC; the rectangle DB, BE is equal to the rectangle AC, BF.

D

E

F

Draw CBG. And since the circumference ABC passes through the centre of AGD, .. (ii. 60.) AB is equal to BG, and the rectangle AB, BC is equal to the rectangle GB, BC, i. e. to the rectangle DB, BE. Also the rectangle AB, BC is equal to the rectangle AC, BF, (Eucl. vi. C.), .. the rectangle DB, BE is equal to the rectangle AC, BF.